`underset("n digits")((666…6)^(2)) + underset("n digits")((888…8))` is equal to

Options

A. `(4)/(9) (10^(n) - 1)`
B. `(4)/(9) (10^(2n) - 1)`
C. `(4)/(9) (10^(n) - 1)^(2)`
D. none of these

Answer

To solve the problem \( (666...6)^2 + (888...8) \) where \( 666...6 \) has \( n \) digits and \( 888...8 \) has \( n \) digits, we will follow these steps:

### Step 1: Express the numbers in a mathematical form
The number \( 666...6 \) with \( n \) digits can be expressed as:
\[ 666...6 = 6 \times (10^{n-1} + 10^{n-2} + \ldots + 10^0) = 6 \times \frac{10^n - 1}{9} \]
Similarly, the number \( 888...8 \) with \( n \) digits can be expressed as:
\[ 888...8 = 8 \times (10^{n-1} + 10^{n-2} + \ldots + 10^0) = 8 \times \frac{10^n - 1}{9} \]

### Step 2: Square the first term
Now, we need to square \( 666...6 \):
\[ (666...6)^2 = \left(6 \times \frac{10^n - 1}{9}\right)^2 = 36 \times \left(\frac{10^n - 1}{9}\right)^2 = \frac{36(10^n - 1)^2}{81} = \frac{4(10^n - 1)^2}{9} \]

### Step 3: Add the second term
Now, we add the second term \( 888...8 \):
\[ (888...8) = 8 \times \frac{10^n - 1}{9} \]

### Step 4: Combine the two results
Now, we combine both results:
\[ (666...6)^2 + (888...8) = \frac{4(10^n - 1)^2}{9} + 8 \times \frac{10^n - 1}{9} \]
Combine the fractions:
\[ = \frac{4(10^n - 1)^2 + 72(10^n - 1)}{9} \]

### Step 5: Simplify the expression
Let \( x = 10^n - 1 \):
\[ = \frac{4x^2 + 72x}{9} \]
Factoring out \( 4 \):
\[ = \frac{4(x^2 + 18x)}{9} \]
Now substitute back \( x = 10^n - 1 \):
\[ = \frac{4((10^n - 1)^2 + 18(10^n - 1))}{9} \]

### Final Result
Thus, the final answer is:
\[ \frac{4(10^{2n} - 2 \cdot 10^n + 1 + 18 \cdot 10^n - 18)}{9} = \frac{4(10^{2n} + 16 \cdot 10^n - 17)}{9} \]